@Chandrasekhar: When we try to go from $ g'(x) + g(x) \geq $ to $g'(x)/g(x) \geq 1 $ we run into the issue of whether $g$ is ever zero, or is negative, flipping the inequality.
–
Ragib ZamanFeb 8 '12 at 10:23

Lierre's proof is essentially the same as mine. The two proofs were posted within one minute.
–
Christian BlatterFeb 8 '12 at 15:12

I wonder if there are any extra assumptions (not given in the original problem) about $y''$ (or $u(x)$) which are being used to solve the differential equation?
–
AryabhataFeb 8 '12 at 16:42

5

Why should $\sinh(x-t) u(t)$ be integrable? It does seem like there might be assumptions being made which aren't specified (or the proof is incomplete). For instance, there are functions which are differentiable at every point, but the derivative is not integrable. $f'(x)$ could be one such function...
–
AryabhataFeb 8 '12 at 19:15

@Aryabhata: one can argue using Darboux sums, see my second answer for details. Basically because the Green's function has a sign, one can compare against the lower Darboux sum and replace the inequality by an inequality with the correct sign.
–
Willie WongFeb 10 '12 at 12:18

The basic intuition is that in a neighbourhood of $0$, $f$ is positive. This leads to $f'$ being increasing (as $f'' > f$), and thus $f$ is increasing. If you picked the "leftmost" point where $f$ is 0, then in that interval, you need $f'$ to become $0$ at some point.

To formalize this, an elementary proof:

Since $f(0) = 0$, and $f'(0) \gt 0$, there is a $\delta \gt 0$ such $f(x) \gt 0$ for all $x \in (0, \delta)$.

This we can see by using the $\epsilon-\delta$ definition of derivative and choosing $\epsilon = \frac{f'(0)}{2}$.

Now assume there is some point $y \gt 0$ where $f(y) \le 0$. This implies there is some point $y' \gt 0$ such that $f(y') = 0$

Now for any $0 \lt x \lt c$, we have that $f(x) \gt 0$. This is because, for $x \lt c$, we cannot have $f(x) = 0$ (as $c = \inf S$), and if $f(d) \lt 0$ for some $d$, then by continuity, there is a point $e \lt d \lt c$ such that $f(e) = 0$.

Thus for $x \in (0,c)$, we have $f(x) \gt 0$ and $f(0) = f(c) = 0$.

In this interval $f''(x) \ge f(x) \gt 0$. Thus $f'(x)$ is increasing, and since $f'(0) \gt 0$, we have that $f'(x) \gt 0$ for all $x \in (0,c)$.

But since $f(0) = f(c)$, we must have that $f'(\eta) = 0$ for some $\eta \in (0,c)$, by Rolle's theorem.

This is a solution taken from the book Berkeley Problems in Mathematics :third edtion by Paulo Ney De Souza and Jorge-Nuno Silva.

Suppose to the contrary that $f(x) \leqslant 0$ for some positive value $x$. Then $a=\inf\{x>0 : f(x)\leqslant 0\}$ is positive. Since $f$ is continuous, $f(a)=0.$ Let $0<b<a$, then $f(b)>0$. So by the mean value theorem, $\exists ~c$ in $(b,a)$ such that $$ f'(c)=\frac{f(a)-f(b)}{a-b} <0.$$
Again, by the mean value theorem, $\exists~ d$ in $(0,c)$ such that $$f''(d)=\frac{f'(c)-f'(0)}{c} <0.$$
But then, $0<d<a$ and $f(d) >0$. However, this contradicts the supposition that $f''(x)\geqslant f(x)$ $\forall~ x$ in $(0,a)$.

Don't we need a proof that $a \gt 0$? Otherwise this is essentially the same as the proof I have.
–
AryabhataFeb 8 '12 at 10:38

@Aryabhata: that follows from continuity of $f'$ (since $f$ is twice differentiable) and the assumption that $f'(0) > 0$. I think that's the "easy" part of the problem. But yeah, it is the same proof as the one you gave.
–
Willie WongFeb 8 '12 at 11:23

@WillieWong: I wasn't claiming it is difficult. Was just being pedantic, and a bit surprised that this was missing from the book.
–
AryabhataFeb 8 '12 at 16:24

+1: for the book. Seems to have some nice problems there.
–
AryabhataFeb 10 '12 at 19:45

Since you asked for a general method, here it is ! And the result proved is much stronger.

Let $g$ denote the function $f''-f$, and consider the differential equation
$$ y''-y=g $$
Obviously $f$ is a solution of this equation.
Assuming that $g$ is locally integrable and using the variation of constants method, one can easily compute the general form of the solutions :
$$ f(t) = C_1 \sinh(t) + C_2 \cosh(t) + \sinh(t)\int_0^tg(x)\cosh(x)dx- \cosh(t)\int_0^tg(x)\sinh(x)dx$$
with $C_1$ and $C_2$ constants.

Evaluating this general solution and its derivative at zero shows that $C_2 = f'(0)$ and $C_2=0$.

Look also at Christian Blatter's answer : it is essentially the same but he cleverly used the addition formula $sh(t-x)=sh(t)ch(x)-sh(x)ch(t)$.
–
LierreFeb 8 '12 at 16:09

Same objection to this as to Christian's answer...
–
AryabhataFeb 10 '12 at 8:28

Yes, and I think this is a good objection. I'm wondering if the integrability is not automatic, considering that $g$ (or $u$ in Christian's answer) is positive.
–
LierreFeb 10 '12 at 9:53

@Lierre: no, that's not automatic. Since $f$ itself is $C^1$, it is integrable. So $g$'s a priori integrability is identical to that of $f''$. Now consider the Volterra function. Some modification to that should give a counterexample (add to it a suitably fast growing smooth function).
–
Willie WongFeb 10 '12 at 12:36

@Willie: To my knowledge the Volterra function is Lebesgue-integrable, isn't it ? Is the derivative of a differentiable monotonic function always (Lebesgue)-integrable ? That would solve the problem.
–
LierreFeb 10 '12 at 14:02

A bit about the abstract nonsense behind the arguments (since there have been several proofs already, and since you asked about "general technique").

Most proofs of statements of this type relies on some type of induction argument. The most basic form is the principle of mathematical induction: let $P\subseteq \mathbb{N}$ be a subset satisfying that $1\in P$ and $p\in P\implies p+1 \in P$, then $P = \mathbb{N}$.

This can be extended to the principle of transfinite induction: let $(X,<)$ be a well ordered set. Let $P\subseteq X$ be a set with the property that
$$ \beta\in P \forall \beta < \alpha \implies \alpha \in P$$
then $P = X$. (A quick proof: suppose not. Then $X\setminus P$ is nonempty, hence since $X$ is well-ordered $\min X\setminus P$ exists. But that element satisfies $\beta < \min X\setminus P \implies \beta\not\in X\setminus P \implies \beta\in P$. This gives a contradiction. Note that trivially $\min X$ is in $P$ since the set of its predecessors is empty, and hence the induction hypothesis is vacuously true.)

Most sets we play with in real analysis, however, are not well-ordered (by the natural ordering). So we need to use some additional structure. A commonly used one is

Continuity argument
Let $(X,\tau)$ be a connected topological space (where $\tau$ is the topology). Then if $P\subseteq X$ satisfies

$P$ is nonempty

$P$ is open

$P$ is closed

then $P = X$.

Applied to your case, you have that $X = \mathbb{R}_+$ with the usual topology. Let $P := \{ x: f(x)>0\}$. Since $f$ is differentiable, it is continuous, and so $P := f^{-1}(\mathbb{R}_+)$ must be open. Using that $f'(0) > 0$ you have that $P$ is nonempty. Hence it remains to show that $P$ is closed. (To show directly that $P$ is closed is hard; what if $x_0$ is a limit point of $P$ approachable only from above? The actual proof given by Aryabhata and Nana via the intermediate value theorem is slightly different (and incorporates a refinement): noting that there exists some $a\in P$ such that $(0,a)\subset P$, in this case it is better to instead consider $P_0$, the connected component of $P$ containing $(0,a)$. Again we have that $P_0$ is open. So in the proof it suffices to show that $P_0$ is closed using the intermediate value theorem.)

(Note: the proof given by Aryabhata and Nana can also be phrased as an induction procedure based on order theory using that the topology and order of the real numbers are compatible, and where $P_0$ is naturally an initial segment, but I'll not go into details there: the basic idea, however, is still the same.)

Therefore define $S_L(y,0;v)$ to be the lower Darboux sum of $v$ between $0$ and $y$, and similarly $S_U(y,0;v)$ to be the upper Darboux sum. We have that $S_L(y,0;y') \leq u(y) - u(0) \leq S_U(y,0;u')$ whenever $u$ is differentiable.

Using that Darboux sums are monotone (meaning that if $u \leq v$ then $S_L(y,0;u) \leq S_L(y,0;v)$ and similarly for the upper sum), we get the following result: if $f$ and $h$ are twice differentiable functions such that $f'' - f \geq h'' - h$ pointwise everywhere with $f'(0) \geq h'(0)$, then we have
$$ f(x) - f(0) \geq \mathcal{G}_L(x,0;h'' -h,h') $$
and
$$ h(x) - h(0) \leq \mathcal{G}_U(x,0;f'' -f,f') $$
In particular, if $h$ is twice continuously differentiable (or simply if $h'' - h$ is Riemann integrable), then the Darboux sums converge to the Riemann integral and which using the fundamental theorem of calculus we get that necessarily
$$ f(x) - f(0) \geq h(x) - h(0)~. $$

The fact about comparing upper and lower Darboux sums against the "primitive" is standard. The key to being able to apply this result uses the fact that the linear differential equation can be formally solved by integrating factors, which is related to the fact that the ODE has signed Green's function. (In the study of linear partial differential equations, the fact that the Green's function has a sign is closely connected to the existence of some form of maximum principle. )

+1: More questions though: Is integrability of $f''$ enough to prove that all the solutions of the differential equation are given by Christian's answer? Are you sure there are no extra assumptions being made there?
–
AryabhataFeb 10 '12 at 18:30

@Aryabhata: defining the function $u := y'' - y$ then it is a tautology that $y'' - y = u$. Whether there exists other solutions is irrelevant. If $u$ is integrable, then by a suitable application of the fundamental theorem of calculus you must have that the representation formula for $y$ hold. (BTW, that a weak version of a theorem is stated on Wikipedia does not preclude a stronger version being true [especially in specific cases].) And lastly, boundary value problems could be a bit different from initial value problems.
–
Willie WongFeb 10 '12 at 19:13

One of the important things to note is the following: if you try to solve $y'' = f$ where $f$ is only locally integrable in the Lebesgue sense, then you only get that $y'$ is absolutely continuous and hence almost everywhere differentiable, and $y$ itself is continuously differentiable with absolutely continuous derivative. In particular, you cannot guarantee twice differentiability. But the second form of the fundamental theorem of calculus tells you that if $y$ is twice differentiable and $y''= f$ is integrable, then any solution of $u'' = f$ is a continuous function which...
–
Willie WongFeb 10 '12 at 19:19

agrees with $y$ everywhere; and hence by our a priori assumption that $y$ is twice differentiable, any solution must also be differentiable. (And in fact $y$ is the only solution.) So yes, there are cases where continuity of the "right hand side" does matter, but this case here is not one of them.
–
Willie WongFeb 10 '12 at 19:21

Integrating both sides of the inequality between $[0,t]$ gives $$ f'(t)-f'(0) \geq \int^t_0 f(x) dx. $$ Suppose there exists positive $y$ such that $f(y)\leq 0.$ There first two conditions imply that there exists some $t_0>0$ where $ f(t) > 0 $ holds for all $ t \in (0,t_0).$ Let $t_1$ be the supremum of all such $t_0.$ Since $f$ is positive but then goes back to $0$, $f'(t_2) < 0 $ for some $ t_2 \in (0,t_1) $. But this contradicts $$ f'(t_2) \geq f'(0) + \int^{t_2}_0 f(x) dx \geq 0. $$